BZOJ 4567: [Scoi2016]背单词
二次联通门 : BZOJ 4567: [Scoi2016]背单词
/* BZOJ 4567: [Scoi2016]背单词 Trie + 贪心 大家都吐槽题目反人类 可我觉得还好,毕竟见的多了 不会做啊。。 正解好巧妙 考虑一下,发现一操作完全不必要,可以省去 因为所有的字符串的后缀关系会形成一个树 那么把字符串倒序插入Trie中 建树,每次向子树小的一个点转移即可 */ #include <cstdio> #include <algorithm> #include <cstring> #define Max 100008 #define L 26 int N; char line[Max]; int trie[Max][L], T_C = 1; bool is[Max]; struct E { E *n; int to; }; E *list[Max], poor[Max], *Ta = poor; void Dfs (int now, int Father) { if (is[now]) { ++ Ta, Ta->to = now, Ta->n = list[Father], list[Father] = Ta; Father = now; } for (int i = 0; i < L; ++ i) if (trie[now][i]) Dfs (trie[now][i], Father); } int size[Max]; void Re_Dfs (int now) { size[now] = 1; for (E *e = list[now]; e; e = e->n) Re_Dfs (e->to), size[now] += size[e->to]; } int dfn[Max], top, C, Stack[Max], Answer; inline bool Comp (int a, int b) { return size[a] < size[b]; } void Get_Answer (int now) { dfn[now] = ++ C; int pos = top + 1; for (E *e = list[now]; e; e = e->n) Stack[++ top] = e->to; if (top < pos) return ; std :: sort (Stack + pos, Stack + top + 1, Comp); for (int i = pos; i <= top; ++ i) Get_Answer (Stack[i]), Answer += dfn[Stack[i]] - dfn[now]; top = pos - 1; } int Main () { scanf ("%d", &N); register int i, j; int Id, x, now, Len; for (i = 1; i <= N; ++ i) { scanf ("%s", line); Len = strlen (line); for (j = Len - 1, now = 1; j >= 0; -- j) { Id = line[j] - 'a'; if (trie[now][Id] == 0) trie[now][Id] = ++ T_C; now = trie[now][Id]; } is[now] = true; } Dfs (1, 1), Re_Dfs (1), Get_Answer (1); printf ("%d", Answer); return 0; } int ZlycerQan = Main (); int main (int argc, char *argv[]) {;}
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